A rectangle PQRS where point A is on PQ such that PA = 2/3 PQ . AS intersects PR at B . It is given that SP = 4 q , SR = 6p , |p| = 3 units and |q| = 2 units . Find vector PB and BA .
I just noticed now that these are supposed to solved using vectors.
I know vectors. The old way. Even if I studied the new or modern way in dealing with vectors, my Math brain rejected this new way.
Just like using matrices in solving simultaneous equations, my stubborn Math brain rejected the matrices.
(Those two ways are making easy Math hard.)
So, I will show a solution not using the vector way. It's just through simple Geometry, and/or trig.
In triangle PBA:
Referring to the right triangle PQR,
tanP = QR /PQ = 4(2) /6(3) = 8/18
angle P = arctan(8/18) = 23.9625 degrees
Referring to right triangle APS,
tanA = 8 / (2/3)18 = 8/12
angle A = arctan(8/12) = 33.69 deg
So, angle B = 180 -23.9625 -33.69 = 122.3475 deg
By Law of Sines,
AP /sin(122.3475deg) = PB /sin(33.69deg)
12 /sin(122.3475deg) = PB /sin(33.69deg)
PB = 7.879 units ----------answer.
By Law of Sines again
12 /sin(122.3475deg) = AB /sin(23.9625deg)
AB = 5.769 units ----------answer.